Problem: What is the average value of $\sin(x)$ on the interval $-6\leq x \leq -1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\cos(6)+\cos(1)}{5}$ (Choice B) B $\dfrac{\cos(6)-\cos(1)}{5}$ (Choice C) C $-\dfrac{\cos(6)+\cos(1)}{7}$ (Choice D) D $-\dfrac{\cos(6)-\cos(1)}{7}$
Explanation: In general, this is the average value of function $f$ over the interval $[a,b]$ : $\dfrac{\int_a^b f(x)\,dx}{b-a}$ In our case, ${f(x)=\sin(x)}$, ${a=-6}$ and ${b=-1}$ : $\begin{aligned} \dfrac{\int_{ a}^{ b} {f(x)}\,dx}{ b- a}&=\dfrac{\int_{{-6}}^{ {-1}} ({\sin(x)})\,dx}{{-1}-{(-6)}} \\\\ &=\dfrac{\Big[-\cos(x)\Big]_{-6}^{-1}}{5} \\\\ &=\dfrac{-\cos(-1)-(-\cos(-6))}{5} \\\\ &=\dfrac{\cos(6)-\cos(1)}{5} \end{aligned}$ In conclusion, the average value of $\sin(x)$ on the interval $-6\leq x \leq -1$ is $\dfrac{\cos(6)-\cos(1)}{5}$.